Question: $\dfrac{ -7e - 5f }{ -3 } = \dfrac{ 9e - 2g }{ 6 }$ Solve for $e$.
Multiply both sides by the left denominator. $\dfrac{ -7e - 5f }{ -{3} } = \dfrac{ 9e - 2g }{ 6 }$ $-{3} \cdot \dfrac{ -7e - 5f }{ -{3} } = -{3} \cdot \dfrac{ 9e - 2g }{ 6 }$ $-7e - 5f = -{3} \cdot \dfrac { 9e - 2g }{ 6 }$ Multiply both sides by the right denominator. $-7e - 5f = -3 \cdot \dfrac{ 9e - 2g }{ {6} }$ ${6} \cdot \left( -7e - 5f \right) = {6} \cdot -3 \cdot \dfrac{ 9e - 2g }{ {6} }$ ${6} \cdot \left( -7e - 5f \right) = -3 \cdot \left( 9e - 2g \right)$ Distribute both sides ${6} \cdot \left( -7e - 5f \right) = -{3} \cdot \left( 9e - 2g \right)$ $-{42}e - {30}f = -{27}e + {6}g$ Combine $e$ terms on the left. $-{42e} - 30f = -{27e} + 6g$ $-{15e} - 30f = 6g$ Move the $f$ term to the right. $-15e - {30f} = 6g$ $-15e = 6g + {30f}$ Isolate $e$ by dividing both sides by its coefficient. $-{15}e = 6g + 30f$ $e = \dfrac{ 6g + 30f }{ -{15} }$ All of these terms are divisible by $3$ Divide by the common factor and swap signs so the denominator isn't negative. $e = \dfrac{ -{2}g - {10}f }{ {5} }$